 # Graph Algorithms Using Depth First Search Prepared by John Reif, Ph.D. Distinguished Professor of Computer Science Duke University Analysis of Algorithms.

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Graph Algorithms Using Depth First Search Prepared by John Reif, Ph.D. Distinguished Professor of Computer Science Duke University Analysis of Algorithms Week 8, Lecture 1

Graph Algorithms Using Depth First Search a)Graph Definitions b)DFS of Graphs c)Biconnected Components d)DFS of Digraphs e)Strongly Connected Components

Readings on Graph Algorithms Using Depth First Search Reading Selection: –CLR, Chapter 22

Graph Terminology GraphG=(V,E) Vertex setV Edge setE pairs of vertices which are adjacent

Directed and Undirected Graphs G directed G undirected uv uv

Proper Graphs and Subgraphs Proper graph: –No loops –No multi-edges SubgraphG ‘ of G G ‘ = (V ‘, E ‘ ) where V ‘ is a subset of V, E ‘ is a subset of E

Paths in Graphs Path p P is a sequence of vertices v 0, v 1, …, v k where for i=1,…k, v i-1 is adjacent to v i Equivalently, p is a sequence of edges e 1, …, e k where for i = 2,…k edges e i-1, e i share a vertex V1V1 V2V2 V k-1 VkVk ekek e2e2 e1e1 VoVo

Simple Paths and Cycles Simple path no edge or vertex repeated, except possibly v o = v k Cycle a path p with v o = v k where k>1 V1V1 V2V2 V k-1 V k = V o

A Connected Undirected Graph

Connected Components of an Undirected Graph

Biconnected Undirected Graphs (or G is single edge)

Size of a Graph Graph G = (V,E) n = |V| = # vertices m = |E| = # edges size of G is n+m 4 2 3 1

Definition of an Undirected Tree Tree T is graph with unique simple path between every pair of vertices n = # vertices n-1 = # edges Forest –set of trees

Definition of a Directed Tree Directed Tree T is digraph with distinguished vertex root r such that each vertex reachable from r by a unique path Family Relationships: - ancestors - descendants - parent - child - siblings r leaves have no proper descendants

An Ordered Tree Ordered Tree –is a directed tree with siblings ordered B A C D E F HI

Preorder Tree Traversal PreorderA,B,C,D,E,F,H,I  root (order vertices as pushed on stack)  preorder left subtree  preorder right subtree B A C D E F HI

Postorder Tree Traversal PostorderB,E,D,H,I,F,C,A  postorder left subtree  postorder right subtree  root (order vertices as popped off stack) B A C D E F HI

Spanning Tree and Forest of a Graph T is a spanning tree of graph G if (1) T is a directed tree with the same vertex set as G (2) each edge of T is a directed version of an edge of G Spanning Forest: forest of spanning trees of connected components of G

Example Spanning Tree of a Graph 1 6 2 4 3 5 7 8 root tree edge back edge 9 1011 12 cross edge

Classification of Edges of G with Spanning Tree T An edge (u,v) of T is tree edge An edge (u,v) of G-T is back edge if u is a descendent or ancestor of v. Else (u,v) is a cross edge

Tarjan ’ s Depth First Search Algorithm We assume a RAM machine model Algorithm Depth First Search

Recursive DFS Procedure

Time Cost of Depth First Search Algorithm on a RAM Input size n = |V|, m = |E| Theorem Depth First Search takes total time cost O(n+m) Proof can associate with each edge and vertex a constant number of operations.

Classification of Edges of G via DFS Spanning Tree T 1 6 2 4 3 5 7 8 G 1 6 2 4 3 5 7 8 T

Edge notation induced by Depth First Search Tree T

Classification of Edges of G via DFS Spanning Tree T (cont ’ d) Note DFS tree T has no cross edges will number vertices by order visited in DFS (preorder of T)

Verifying Vertices are Descendants via Preordering of Tree Suppose we preorder number a Tree T Let D v = # of descendants of v Lemma u is descendant of v iff v < u < v + D v v u DvDv W *

Testing for Proper Ancestors via Preordering Lemma If u is descendant of v and (u,w) is back edge s.t. w < v then w is a proper ancestor of v

Low Values For each vertex v, define low(v) = min ( {v} ∪ {w | v→ - - - w} ) Can prove by induction: Lemma can be computed during DFS in postorder. *

Graph with DFS Numbering and Low Values in Brackets 7 1 2 4 3 5 8 6 v[low(v)] Number vertices by DFS number

Biconnected Components G is Biconnected iff either (1) G is a single edge, or (2) for each triple of vertices u,v,w

Example of Biconnected Components of Graph G Maximal subgraphs of G which are biconnected. 5 1 8 1 6 2 4 3 5 7 8 1 2 2 3 4 6 5 7  G

Biconnected Components Meet at Articulation Points The intersection of two biconnected components consists of at most one vertex called an Articulation Point. Example: 1,2,5 are articulation points 1 6 2 4 3 5 7 8 G

Discovery of Biconnected Components via Articulation Points –If can find articulation points then can compute biconnected components: Algorithm: During DFS, use auxiliary stack to store visited edges. Each time we complete the DFS of a tree child of an articulation point, pop all stacked edges currently in stack These popped off edges form a biconnected component

Characterization of an Articulation Point Theorem a is an articulation point iff either (1) a is root with ≥ 2 tree children or (2) a is not root but a has a tree child v with low (v) ≥ a (note easy to check given low computation)

Proof of Characterization of an Articulation Point Proof The conditions are sufficient since any a-avoiding path from v remains in the subtree T v rooted at v, if v is a child of a To show condition necessary, assume a is an articulation point.

Characterization of an Articulation Point (cont ’ d) Case (1) If a is a root and is articulation point, a must have ≥ 2 tree edges to two distinct biconnected components. Case(2) If a is not root, consider graph G - {a} which must have a connected component C consisting of only descendants of a, and with no backedge from C to an ancestor of v. Hence a has a tree child v in C and low (v) ≥ a

Proof of Characterization of an Articulation Point (cont ’ d) a v root Articulation Point a is not the root subtree T v C no back edges Case (2)

Computing Biconnected Components Theorem The Biconnected Components of G = (V,E) can be computed in time O(|V|+|E|) using a RAM Proof idea Use characterization of Bicoconnected components via articulation points Identify these articulation points dynamically during depth first search Use a secondary stack to store the edges of the biconnected components as they are visited When an articulation point is discovered, pop the edges of this stack off to output a biconnected component

Biconnected Components Algorithm  initialize a STACK to empty During a DFS traversal do  add visited edge to STACK  compute low of visited vertex v using Lemma  test if v is an articulation point  if so, for each u ∈ children (v) in order where low (u) > v do Pop all edges in STACK upto and including tree edge (v,u) Output popped edges as a biconnected component of G od

Time Bounds of Biconnected Components Algorithm Time Bounds: Each edge and vertex can be associated with 0(1) operations. So time O(|V|+|E|).

Depth First Search in a Directed Graph Depth first search tree T of Directed Graph cross cycle forward cycle 1 1 2 3 4 5 6 7 8 2 4 5 6 7 8 G = (V,E) i = DFS number v i i = postorder = order vertex popped off DFS stack edge set E partitioned

Classification of Directed Edge (u, v) via DFS Spanning Tree T Tree edge Cycle edge Forward edge Cross edge

Topological Order of Acyclic Directed Graph Digraph G = (V,E) is acyclic if it has no cycles

Characterization of an Acyclic Digraph Proof Case (1): Suppose (u,v) ∈ E is a cycle edge, so v → u. But let e 1,…e k be the tree edges from v to u. Then (u,v), e 1,…e k is a cycle. *

Characterization of an Acyclic Digraph (cont ’ d) Case (2): Suppose G has no a cycle edge Then order vertices in postorder of DFS spanning forest (i.e. in order vertices are popped off DFS stack). This is a reverse topological order of G. So, G can have no cycles. Note: Gives an O(|V|+|E|) algorithm for computing Topological Ordering of an acyclic graph G = (V,E)

Strong Components of Directed Graph Strong Component Collapsed Graph G* derived by collapsing each strong component into a single vertex. note: G* is acyclic.

Directed Graph with Strong Components Circled Directed Graph G = (V,E) 3 1 2 4 7 5 6 8

Algorithm Strong Components Input digraph G  Perform DFS on G. Renumber vertices by postorder.  Let G - be the reverse digraph derived from G by reversing direction of each edge.

Algorithm Strong Components (cont ’ d)  Perform DFS on G -, starting at highest numbered vertex. Output resulting DFS tree of G - as a strongly connected component of G.  repeat , starting at highest numbered vertex not so for visited (halt when all vertices visited)

Time Bounds of Algorithm Strong Components Time Bounds each DFS costs time c(|V|+|E|) So total time = 2*c(|V|+|E|) ≤ O(|V|+|E|)

Example of Step  of Strong Components Algorithm 3 1 2 4 7 5 6 8 8 7 6 5 3 4 2 1 Example Input Digraph G

Example of Step  of Strong Components Algorithm 3 1 2 4 7 5 6 8 8 7 6 5 3 4 2 1 Reverse Digraph G -

Proof of Strong Components Algorithm Theorem The Algorithm outputs the strong components of G. Proof We must show these are exactly the vertices in each DFS spanning forest of G -

Proof of Strong Components Algorithm (cont ’ d) Suppose: v,w in the same strong component and DFS search in G - starts at vertex r and reaches v. Then w will also be reached. So v,w are output together in same spanning tree of G -.

Proof of Strong Components Algorithm (cont ’ d) Suppose: v,w output in same spanning tree of G -. Let r be the root of that spanning tree. Then there exists paths in G - from r to each of v and w. So there exists paths in G to r from each of v and w.

Proof of Strong Components Algorithm (cont ’ d) Suppose: no path in G to r from v. Then since r has a higher postorder than v, there is not path in G from v to r, a contradiction. Hence, there exists path in G from r to v, and similar argument gives path from r to w. So v and w are in a cycle of G and must be in the same strong component.

Graph Algorithms Using Depth First Search Prepared by John Reif, Ph.D. Distinguished Professor of Computer Science Duke University Analysis of Algorithms Week 8, Lecture 1

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